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\[\LARGE xe^{\frac{x^2}{2}}\] this one?

do you know u-substitution?

so what is the derivative of \[\large \frac{1}{2}x^2 ? \]

so what is the derivative of f(x) = 1/2 ?

similarly what is the derivative of \[\large g(x) = x^2 \]

so if \[u= (1/2)x^2 -> u' = x\]

yeah but we have to write this differently
it will be
du = x dx

so now we have \[xe^u dx\]

ok, so \[xe^u dx -> xe^u (1/x) dy -> e^u du\] (+c...)

du

you don't have to write dx
you don't have to write du

Ok, I think I understand, thank you very much!

but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation

and once we have our antiderivative, we can simply check by taking the derivative.

you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]

now take the derivative of that equation...

that's how to check :)

so we get back to \[e^((x^2)/2) \]

yes, +c

mhm

if you take the derivative of your antiderivative, we should be getting that x in front of that e

Yes, I wrote the integral

the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

Excellent. Thanks a lot!